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A right circular cylinder, open at the top is filled with liquid, sp. gr. 1.2 and rotated about its vertical axis at such speed that half the liquid spills out. The pressure at the centre of the bottom is
- a)zero
- b)one-fourth its value when cylinder was full
- c)one-half its value when cylinder was full
- d)same as before rotation
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A right circular cylinder, open at the top is filled with liquid, sp. ...
After half the water has spilled out the vertex of the paraboloid will touch the bottom of cylinder thus pressure will be zero.
Most Upvoted Answer
A right circular cylinder, open at the top is filled with liquid, sp. ...
Explanation:
When a right circular cylinder is rotated about its vertical axis, the liquid inside the cylinder experiences centrifugal force. This force pushes the liquid towards the outer walls of the cylinder, resulting in a concave shape of the liquid surface.
Key Point 1:
When the cylinder is full, the pressure at the bottom is equal to the pressure due to the weight of the liquid column above it.
Key Point 2:
When the cylinder is rotated such that half the liquid spills out, the remaining liquid forms a concave shape. The pressure at the bottom is now not only due to the weight of the liquid column but also due to the centrifugal force acting on the liquid. This centrifugal force creates an additional pressure at the bottom.
Key Point 3:
However, at the center of the bottom of the cylinder, the centrifugal force is zero. This is because the center of the bottom is at the axis of rotation, where the centrifugal force is directed towards the axis and cancels out. Therefore, the pressure at the center of the bottom is only due to the weight of the liquid column above it.
Key Point 4:
Since half of the liquid spills out, the weight of the liquid column above the center of the bottom is reduced by half. As a result, the pressure at the center of the bottom becomes zero because there is no liquid column above it to exert any pressure.
Therefore, the correct answer is option 'A' - the pressure at the center of the bottom is zero.
When a right circular cylinder is rotated about its vertical axis, the liquid inside the cylinder experiences centrifugal force. This force pushes the liquid towards the outer walls of the cylinder, resulting in a concave shape of the liquid surface.
Key Point 1:
When the cylinder is full, the pressure at the bottom is equal to the pressure due to the weight of the liquid column above it.
Key Point 2:
When the cylinder is rotated such that half the liquid spills out, the remaining liquid forms a concave shape. The pressure at the bottom is now not only due to the weight of the liquid column but also due to the centrifugal force acting on the liquid. This centrifugal force creates an additional pressure at the bottom.
Key Point 3:
However, at the center of the bottom of the cylinder, the centrifugal force is zero. This is because the center of the bottom is at the axis of rotation, where the centrifugal force is directed towards the axis and cancels out. Therefore, the pressure at the center of the bottom is only due to the weight of the liquid column above it.
Key Point 4:
Since half of the liquid spills out, the weight of the liquid column above the center of the bottom is reduced by half. As a result, the pressure at the center of the bottom becomes zero because there is no liquid column above it to exert any pressure.
Therefore, the correct answer is option 'A' - the pressure at the center of the bottom is zero.
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A right circular cylinder, open at the top is filled with liquid, sp. gr. 1.2 and rotated about its vertical axis at such speed that half the liquid spills out. The pressure at the centre of the bottom isa)zerob)one-fourth its value when cylinder was fullc)one-half its value when cylinder was fulld)same as before rotationCorrect answer is option 'A'. Can you explain this answer?
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